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Setler79 [48]
3 years ago
8

What is true of the data in the dot plot? Check all that apply.

Mathematics
2 answers:
otez555 [7]3 years ago
8 0

True answers for data in the plot are

The center is 13

The center is not 14

The Peak is 14.

It has three clusters

It is not symmetric to left or right it is bi-modal.

It is has a range from 10 to 15 most number of data points are 13 to 15.

Total number times Shelly waited is 16 times.

Step-by-step explanation:

  • While taking cumulative frequency 56.75 percentage comes in 13%.
  • It is the center point of the data.
  • The 14  is not the center as it shows 93rd percent.
  • It has three clusters 0-2 has one cluster,2-4 has 2nd cluster,5-8 third.
  • It is not skewed on the left is has bi modal frequency has two heights.
  • The person indeed waited for 16 times adding total dots.
  • There was a zero 12 which created bi-modal distribution

bonufazy [111]3 years ago
8 0
<h2>Answer:</h2>

The correct answer is A,G,E

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Step-by-step explanation:

m^2 + n^2

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Answer:

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Step-by-step explanation:

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2 years ago
2(w – 3y + 4) + w – 5
melisa1 [442]

Answer: It should be 3w - 6y + 3

Step-by-step explanation:

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Read 2 more answers
A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student
Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that p = 0.25

Class of 18 students

This means that n = 18

What would be the probability that at least 6 students drive themselves to school?

This is

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096

P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17

P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213

P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282

Closest option is B, just a small rounding difference.

4 0
3 years ago
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